= Solution
The <orthogonal> coordinate change preserves the <kinetic energy> and takes the <particle exchange operator> to $(x_+,x_-)\mapsto(x_+,-x_-)$. Thus the spatial <wave function> is even in $x_-$ for <bosons> and odd for <fermions>.
The total <potential energy> and <Hamiltonian> separate as
$$
V=\frac{m\omega^2}{2}x_+^2+\frac{m\omega^2(1+2\kappa)}2x_-^2,\qquad
H=H_+(\omega)+H_-(\Omega),\quad\Omega=\omega\sqrt{1+2\kappa}.
$$
Assume $1+2\kappa>0$ so the relative <oscillator> is confining. Its spectrum is $E_{n_+,n_-}=\hbar\omega(n_++1/2)+\hbar\Omega(n_-+1/2)$, and its <parity> in the <two-particle relative coordinate> is $(-1)^{n_-}$.
The lowest <boson> state uses $n_+=n_-=0$, whereas the lowest <fermion> state uses $n_+=0$, $n_-=1$. Therefore
$$
\boxed{E_B=\tfrac12\hbar\omega[1+\sqrt{1+2\kappa}],\qquad
E_F=\tfrac12\hbar\omega[1+3\sqrt{1+2\kappa}].}
$$
For an <oscillator> of frequency $\Omega$, $\langle x_-^2\rangle_n=(n+1/2)\hbar/(m\Omega)$. Since $(x_1-x_2)^2=2x_-^2$,
$$
\boxed{\langle(x_1-x_2)^2\rangle_B=\frac\hbar{m\Omega},\qquad
\langle(x_1-x_2)^2\rangle_F=\frac{3\hbar}{m\Omega}.}
$$
The factor three is the extra relative-coordinate excitation required by antisymmetry. The supplied <oscillator> <wave functions> correspond to <kinetic energy> $P^2/(2m)$; the extra factor $1/2$ printed before that term in the hint is inconsistent with those functions and the stated energies. The calculation uses the conventional <kinetic energy> dictated by particle <mass> $m$.
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